If $x$ is a real number and $x^2-7x+6<0$, what are the possible values for $x$? Use interval notation to express your answer.
Solution: Factoring the quadratic gives $(x-1)(x-6)<0$, which means that $x-1$ and $x-6$ must have opposite signs, since the product of two factors with the same sign is positive. Now, we split into four cases. If $x<1$, then both factors are negative. If $x>6$, both factors are positive. If $x=1$ or $x=6$, one of the factors is zero. If $1<x<6$, $x-6$ is negative and $x-1$ is positive, meaning the inequality is satisfied. So our only possible range is $1<x<6$, giving interval notation of $\boxed{(1,6)}$.